show that the quadrilateral formed by joining the midpoints of the consecutive side of a square is also a square
Let ABCD is a square such that AB = BC = CD = DA, AC = BD and P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.
We know that, diagonals of a square are perpendicular bisector of each other.
∴∠AOD = ∠AOB = ∠COD = ∠BOC = 90°
Now, in quadrilateral EHOS, we have
SE || OH, therefore, ∠AOD + ∠AES = 180° [corresponding angles]
⇒ ∠AES = 180° - 90° = 90°
Again, ∠AES + ∠SEO = 180° [linear pair]
⇒ ∠SEO = 180° - 90° = 90°
SH || EO, therefore, ∠AOD + ∠DHS = 180° [corresponding angles]
⇒ ∠DHS = 180° - 90° = 90°
Again, ∠DHS + ∠SHO = 180° [linear pair]
⇒ ∠SHO = 180° - 90° = 90°
Again, in quadrilateral EHOS, we have
∠SEO = ∠SHO = ∠EOH = 90°
Therefore, by angle sum property of quadrilateral in EHOS, we get
∠SEO + ∠SHO + ∠EOH + ∠ESH = 360°
90° + 90° + 90° + ∠ESH = 360°
⇒ ∠ESH = 90°
In the same manner, in quadrilateral EFOP, FGOQ, GHOR, we get
∠HRG = ∠FQG = ∠EPF = 90°
Therefore, in quadrilateral PQRS, we have
PQ = QR = SR = PS and ∠ESH = ∠HRG = ∠FQG = ∠EPF = 90°
Hence, PQRS is a square.